Demonstrations of Dirac equation covariance state:(adsbygoogle = window.adsbygoogle || []).push({});

The Dirac equation is

[itex](i γ^{μ} ∂_{μ} - m)ψ(x) = 0. \ \ \ \ \ \ \ \ \ \ [/itex] [1]

If coordinates change in a way that

[itex]x \rightarrow x' = Lx, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [/itex] [2]

where [itex]L[/itex] is a Lorentz transformation, [1] should mantain its form, obtaining in the new system:

[itex](i γ^{μ} ∂'_{μ} - m)ψ'(x') = 0, \ \ \ \ \ \ \ \ [/itex] [3]

where

[itex]ψ'(x') = S(L)ψ(x) \ \ \ \ \ \ \ \ [/itex] [4]

and [itex]S(L)[/itex] is an invertible matrix rappresenting the fact [itex]ψ'(x')[/itex] should be a linear combination of [itex]ψ(x)[/itex] and should depend on [itex]L[/itex].

Remembering that from [2] stems

[itex]∂ \rightarrow ∂' = L^{-1}∂, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [/itex] [5]

and substituting [4] and [5] in [3] we obtain

[itex](i γ^{μ} L^{-1 ρ}_{μ} ∂_{ρ} - m)Sψ(x) = 0. \ \ \ \ \ \ \ \ \ [/itex] [6]

Multiplying on the left for [itex]S^{-1}[/itex]:

[itex](i S^{-1} γ^{μ} L^{-1 ρ}_{μ} ∂_{ρ}S - m)ψ(x) = 0[/itex]

and because [itex]S[/itex] depends on [itex]L[/itex] that don't vary along coordinates

[itex](i S^{-1} γ^{μ} L^{-1 ρ}_{μ}S∂_{ρ} - m)ψ(x) = 0. \ \ \ \ \ \ \ \ \ [/itex] [7]

To obtain covariance [7] must be equals to [1] so

[itex]S^{-1} γ^{μ} L^{-1 ρ}_{μ}S = γ^{ρ}.\ \ \ \ \ \ \ \ \ [/itex] [8]

At this point all books state that [8] is equivalent to say

[itex]S^{-1} γ^{μ} S = L^{μ}_{ρ}γ^{ρ}.\ \ \ \ \ \ \ \ \ [/itex] [9]

This requires [itex]S[/itex] and [itex]L[/itex] to commute.

I don' understand how it comes. They are both Lorentz transformations so not necessarly commute.

Thanks

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# Demonstration of Dirac equation covariance

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